Examples percent composition

The percent composition of a compound is specified by citing the mass percents of the elements present. For example, in a 100-g sample of water there are 11.19 g of hydrogen and 88.81 g of oxygen. Hence the percentages of the two elements are... 

Use the formula of a compound to find percent composition or its equivalent (Example 3.4 Problems 33-42) 34,36... 

The elemental analysis of a compound is usually determined by a laboratory that specializes in this technique. A chemist who has prepared a new compound sends a sample to the laboratory for analysis. The laboratory charges a fee that depends on the type and number of elements analyzed. The results are returned to the chemist as a listing of mass percent composition. The chemist must then figure out which chemical formula matches this composition. If a chemist has reason to expect a particular chemical formula, the observed percentages can be matched against the calculated percentages for the expected formula. This process is illustrated in Example 3-13. 

In the past, ionomers have generally consisted of 10-12 mole percent of ions and it is our intention to be consistent with the corresponding random ionomers previously discussed in the literature. In addition to gel permeation chromatography (GPC), H and 3C NMR can readily be utilized to verify the relative amount of monomer successfully incorporated into the block copolymer. For example, the composition of a PMMA-PTBMA diblock can be verified by H NMR ratioing the methyl ester integration (3.5 ppm) to the t-butyl ester integration (1.36 ppm). Figure 1 depicts the t-butyl ster chemical shift which appears reproducibly at 1.J6 ppm. C or FTIR can be utilized in certain instances when H NMR chemical shifts overlap. For... 

EXAMPLE 4.10. Calculate the percent composition of MgS04 that is, the percent by mass of each element in the compound. 

If we know the formula of a compound, it is a simple task to determine the percent composition of each element present. For example, suppose you wanted the percentage carbon and hydrogen in methane, CH4. First, calculate the molecular mass of methane ... 

Chemists are often concerned with precisely what percentage of a compound s mass consists of one particular element. Lying awake at night, uttering prayers to Avogadro, they fret over this quantity, called percent composition. Calculating percent composition is trickier than you may think. Consider the following problem, for example. 

The beauty of this little trick is that you conveniently gift yourself with the same number of grams of each elemental component as its contribution to the percent composition. For example, if you assume that you have 100 g of a compound composed of 60.3% magnesium and 39.7% oxygen, you know that you have 60.3 g of magnesium and 39.7 g of oxygen. (The only time you don t do this is if the problem specifically gives you the masses of each element present in the unknown compound.)... 

Because the original percent composition data is typically experimental, expect to see a bit of error in the numbers. For example, 2.03 is probably within experimental error of 2, 2.99 is probably 3, and so on. 

Figure 1. Example of compositionally resolved bimodal and monomodal distributions of aerosols. The ordinate gives the percent of the species found in the given size fraction of the impactor. The mode near 0.3 xm is the accumulation mode , and that above 8 xm is the coarse mode The minimum of mass between 1 and 2 xm is typical the chlorine distribution is anomalous. Chlorine is in fact a coarse-mode marine aerosol that has lost its larger particles during transport from the ocean to Davis, California, a distance of roughly 100 km. (Reproduced with permission from reference 15. Copyright 1988.)...

Whenever a new compound is made in the laboratory or found in nature, it must be analyzed to find what elements it contains and how much of each element is present—that is, to find its composition. The percent composition of a compound is expressed by identifying the elements present and giving the mass percent of each. For example, we might express the percent composition of a certain colorless liquid found in gasoline by saying that it contains 84.1% carbon and 15.9% hydrogen by mass. In other words, a 100.0 g sample of the compound contains 84.1 g of carbon atoms and 15.9 g of hydrogen atoms. 

Just as we can derive the empirical formula of a substance from its percent composition, we can also calculate the percent composition of a substance from its empirical (or molecular) formula. The strategies for the two kinds of calculations are exactly opposite. Aspirin, for example, has the molecular formula C9H8O4 and thus has a CH 0 mole ratio of 9 8 4. We can convert this mole ratio into a mass ratio, and thus into percent composition, by carrying out mole-to-gram conversions. 

Worked Examples 3.15 and 3.16 show further conversions between percent composition and empirical formulas. 

Percent composition, also called percent by mass, is a useful piece of data to obtain when looking at the composition of certain substances. In percent composition problems you are asked to find the percent of the mass of an element in a compound as compared to the molar mass of the compound. A simple ratio will suffice and the result is multiplied by 100%. For CaCl2, for example, what percent of this compound is made up of chlorine The total mass is... 

Sedimentary rocks are typically a result of erosion and the interaction of the environment on igneous and metamorphic rocks they are formed by the aggregation of mineral particles or sediments. Their SiCVCaO percent composition can be quite varied for example, from 74.3/4.9 (sandstones) down to 8.2/40.5 (limestones). On the average, sedimentary... 

The steps we take to obtain an empirical formula from percent composition data are given in the left column (Steps) that follows. In the right column (Example), the empirical formula of a compound containing 39.2% phosphorus and 60.8% sulfur is calculated. 

Formulas describe the composition of compounds. Empirical formulas give the mole ratio of the various elements. However, sometimes different compounds have the same ratio of moles of atoms of the same elements. For example, acetylene, C2H2, and benzene, CeHe, each have 1 1 ratios of moles of carbon atoms to moles of hydrogen atoms. That is, each has an empirical formula CH. Such compounds have the same percent compositions. However, they do not have the same number of atoms in each molecule. The molecular formula is a formula that gives all the information that the empirical formula gives (the mole ratios of the various elements) plus the information of how many atoms are in each molecule. In order to deduce molecular formulas from experimental data, the percent composition and the molar mass are usually determined. The molar mass may be determined experimentally in several ways, one of which will be described in Chap. 12. 

Sgroup Data. In MDL structure storage, the attachment of structure-differentiating data directly to the structure. Such data may relate to the structure as a whole, or to atoms, bonds, fragments, or collections of atoms and bonds. Examples would include atomic partial charges on 3D models or percent composition attached to components of a formulation. 

Large supply of naturally derived lipids can be obtained from plants in which many oils and fatty acids can be readily extracted and purified. Animal sources (e.g., eggs or milkfats) are used to derive complex lipids such as phospholipids and cholesterol. Yield from natural sources is dependent on the weight-percent composition and the efficiency of the extraction procedure. The constitution of fatty acids in vegetable oils varies widely from different sources. For example, oleic acid is present at 64.6% by weight in olive oil but is present at only 0.7% in palm kernel oil. Similarly, castor oil triglyceride is comprised of almost entirely ricinoleic chains. There are numerous raw material suppliers of oils and oil fractions worldwide. As such, the relative cost of bulk purified... 

The data used to determine the chemical formula for a compound may be in the form of percent composition or it may be the actual masses of the elements in a given mass of the compound. If percent composition is given, you can assume that the total mass of the compound is 100.00 g and that the percent by mass of each element is equal to the mass of that element in grams. For example, the percent composition of an oxide of sulfiir is 40.05% S and 59.95% O. Thus, as you can see in Figure 11-10, 100.00 g of the oxide contains 40.05 g S and 59.95 g O. The mass of each element can be converted to a number of moles by multiplying by the inverse of the molar mass. Recall that the number of moles of S and O are calculated in this way. 

To avoid uncertainty, always specify explicitly the type of percent composition being discussed. If this information is missing, the user must decide intuitively which of the several types is involved. The potential error resulting from a wrong choice is considerable. For example, commercial 50% (w/w) sodium hydroxide contains 763 g of the reagent per liter, which corresponds to 76.3% (w/v) sodium hydroxide. 

If the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition). For example, one... 

Science, mathematics, and technology are interconnected. Teaching Chemistry incorporates the other sciences as well as other disciplines, such as mathematics. For example, graphs and charts are frequently used to record and analyze data. On a daily basis, we are surrounded with mathematics in the ability to make various measurements of mass and size, in the conversions betweens the numerous units, and in tabulating amounts of materials. Beyond these basic skills, mathematical and algebraic skills are used in a plethora of chemical calculations from determining the percent composition of elements in a compound to estimating the mass of reactants needed in a reaction. 

Calculating the percent composition of ionic compounds that contain polyatomic ions requires a careful counting of all the atoms of each element. For example, in 1 mole of ammonium phosphate, (NH4)3P04, there are 3 moles of ammonium ions, NH4+, and 1 mole of phosphate ion, P043". In terms of the elements ... 

Here s another example. Determine the empirical formula of an oxide of iron from its percent composition Fe = 69.94% O = 30.06%. 

The procedure used in the example can be reversed if necessary. Given the percent composition by mass of a compound, we can determine the empirical formula of the compound. Since we are dealing with percentages and the sum of aU the percentages is 100 percent, it is convenient to assume that we started with 100 g of a compound, as Example 3.9 shows. 

The fact that we can determine the empirical formula of a compound if we know the percent composition allows us to identify compounds experimentally. The procedure is as follows. First, chemical analysis tells us the number of grams of each element present in a given amount of a compound. Then we convert the qnantities in grams to nnmber of moles of each element. Finally, using the method given in Example 3.9, we find the empirical formnla of the compound. 

The formula calculated from percent composition by mass is always the empirical formula because the coefficients in the formula are always reduced to the smallest whole numbers. To calculate the actual, molecular formula we must know the approximate molar mass of the compound in addition to its empirical formula. Knowing that the molar mass of a compound must be an integral multiple of the molar mass of its empirical formula, we can use the molar mass to find the molecular formula, as Example 3.11 demonstrates. 

The table below lists the percent composition by mass of nitrogen in some common fertilizers. The preparation of urea was discussed in Example 3.15. 

Check the answer to see that it is reasonable. Some problems have reasonable checks built in, like the percent composition problems in Section 4.3. If the percentages don t add up to 100%, there is a mistake somewhere. For others, we can use the answer to calculate one of the original values, as in empirical formula problems (Section 4.4). Still others require that we know the range of possibilities for our answer. For example, if we get a molarity of 10,000 M (Section 6.1) we know there is a mistake, because... 

EXAMPLE 10 Calculate the percent composition of NH4NO3, another fertilizer. 

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